What is the equation of the quadratic graph with a focus of (4, −3) and a directrix of y = −6?
f(x) =negative one over six times x squared plus four over three times x plus eleven over six
f(x) =one over six times x squared minus four over three times x minus eleven over six
f(x) = three fiftieth x squared plus eleven twenty fifth x plus three fifth
f(x) = two fiftieth x squared plus eleven twenty fifth x plus two fifth

- anonymous

- katieb

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- anonymous

- anonymous

- Nnesha

first we should draw a point at (4,-3) and horizontal line y=-6
|dw:1442578792483:dw| this one is better

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## More answers

- anonymous

|dw:1442578889393:dw|

- Nnesha

|dw:1442578826480:dw|
yes that's right :=)

- Nnesha

brb wait a mint plz

- Nnesha

sorry about alright
so now we should pick a point to find distance between the point and focus
and distance between the point and directrix
you can pick any point |dw:1442579344198:dw|

- Nnesha

equation would be like this \[\huge\rm \sqrt{(x-a)^2+(y-b)^2} =\sqrt{ (y-c)^2}\]
c = directrix
(a,b) is a focus point (4,-3) now substitute a ,b c for their values

- anonymous

square root 4-a^2+-3-b=square root-3-c^2

- Nnesha

well (x,y) is a point that we picked we supposed taht (a,b) is the focus point
so replace a with 4 b with -3 and c with -6
c is directix

- anonymous

4-4^2+-3-(-3)=square root-3-(-6)^2 like this?

- Nnesha

keep the x and y variable \[\large\rm \sqrt{(x-4)^2+(y-(-3))^2} =\sqrt{ (y-(-6))^2}\]
like this now take square both sides to cancel out the square root

- anonymous

X^-16+Y+9=Y+36

- anonymous

X^2

- anonymous

is that right

- Nnesha

hmm no (x-4)^2 = ?

- anonymous

X^2+16

- Nnesha

hmm no (x-4)^2 is same as (x-4)(x-4) foil it signs are the same in the parentheses so you should get 3 terms

- anonymous

X^2-8x+16

- Nnesha

yes right \[x^2-8x+16+(y-(-3))^2=(y-(-6))^2\] foil (y-(-3))^2 and then right side

- Nnesha

here is a hint you will get two terms when you have (a-b)(a+b) opposite signs

- anonymous

Y^2-9

- Nnesha

hmm no (y-(-3))^2 first distribute -1(-3) ?

- anonymous

thats 3

- Nnesha

yes so (y+3)^2 is same as (y+3)(y+3)

- anonymous

oh

- anonymous

y^2+6y-9

- anonymous

+9

- Nnesha

yes right \[\large\rm x^2-8x+16+y^2+6y+9=(y-(-6))^2\] foil right side

- anonymous

Y^2+12y+36

- anonymous

i dont mean to rush but i have to go soon

- Nnesha

yes right \[\large\rm x^2-8x+16+y^2+6y+9=y^2+12y+36\] foil right side
now simplify
combine like terms

- Nnesha

well it will takes some time we need to apply the complete the square method .....

- Nnesha

\[\large\rm x^2-8x+\color{red}{16}+\color{blue}{y^2}+6y+\color{red}{9}=\color{blue}{y^2}+12y+36\]
combine like terms at left side
and then move y^2 to the right side let me know what you get

- anonymous

srry but i have to go thank you any ways

- Nnesha

okay we will finish this whnevr u hve time

- princesssleelee

@Nnesha i have this question as well that i need assistance with.

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